1

Chapter 4 Vapor Pressure An important goal of this chapter is to learn techniques to calculate vapor pressures To do this we will need boiling points and entropies of vaporization The saturation vapor pressure is defined as the gas phase pressure in equilibrium with a pure solid or liquid.

• fi = γi X ipiL*

pure liquid

• constplogTSPC

Clog *

iLgas

part +−=

• KiH = p*iL/Ciw

sat

2

Vapor pressure and Temperature

dGliq = dGgas

from the 1st law H= U+PV

dH = dU + Vdp+pdV

dU= dq - dw

for only pdV work, dw = pdV and from the definition of entropy, dq = TdS

dU = TdS –pdV

from the general expression of free energy

dG = dU + Vdp+pdV- SdT-TdS

substituting for dU

dG= +VdP - SdT

3

The molar free energy Gi/ni = µi

for a gas in equilibrium with a liquid

dµliq = dµgas

dµi liq = Vi liqdp - Si liqdT

Vi liqdpi - Si liqdT = Vigasdp - SigasdT

d/dT = (Sigas -Si liq)/Vigas

at equilibrium ∆G = ∆H -∆S T= zero so (Sigas -Si liq) = ∆Hi vap/T

substituting

dpi /dT = ∆Hi /( Vigas T)

(Clapeyron eq)

substituting Vi gas = RT/pi

RH

)T(d)p(lnd;RTxT

pHdTdp i

iiii ∆

−=∆

= 1

4

constTR

Hpln io

i +∆

−=1

5

Figure 4.3 page 61 Schwartzenbach

6

7

This works over a limited temperature range w/o any phase change Over a larger range Antoine’s equation may be used

ACT

Bln +

+=*

ip

over the limits P\p1 to P2 and T1 to T2

log( )

.PP

vapH T T

RT T2

1

2 1

1 22 303=

−∆

If the molar heat of vaporization, ∆Hvap of hexane equals 6896 cal/mol and its boiling point is 69oC, what is its vapor pressure at 60oC

KKKmolcal

KKmolcalP 3423339913032

3333426896760

1 ∗∗∗−=

/..)(/

log

P1= 578 mm Hg

8

Below the melting point a solid vaporizes w/o melting, that is it sublimes

A subcooled liquid is one that exists below its melting point.

• We often use pure liquids as the reference state

• logKp Log p*

i

9

Molecular interaction governing vapor pressure

As intermolecular attractive forces increase in a liquid, vapor pressures tend to decrease

van der Waals forces generally enthalpies of vaporization increase with increasing polarity of the molecule Both boiling points and entropies of vaporization become important parameters in estimating vapor pressures

10

A constant entropy of vaporization Troutons rule Figure 4.5; at 25oC

This suggests that ∆vapS may tend to be constant

11

At the boiling point is ∆∆vapSTb constant? ∆H const slope = ∆S T Tb

oC ∆vapH ∆vapS kJ mol-1 kJ mol-1K-

1

n-hexane 68.7 28.9 n-decane 174.1 38.8 ethanol 78.3 38.6 naphthalene 218 43.7 Phenanthrene 339 53.0 Benzene 80.1 30.7 Chlorobenzene 131.7 35.2 Hydroxybenzene 181.8 45.7

12

Predicting ∆∆vapSTb

Kistiakowsky derived an expression for the entropy of vaporization which takes into account van der Waal forces

• ∆∆Svap= 36.6 +8.31 ln Tb (eq 4-20)

• for polarity interactions Fistine proposed ∆∆Svap= Kf (36.6 +8.31 ln Tb) Kf= 1.04; esters, ketones Kf= 1.1; amines Kf= 1.15; phenols Kf= 1.3; aliphatic alcohols

Calculating ∆∆Svap using chain flexibility and functionality (Mydral et al, 1996) ∆∆vapSi (Tb) = 86.0+ 0.04 ττ + 1421 HBN (eq 4-21)

τ = Σ(SP3 +0.5 SP2 +0.5 ring) -1

13

SP3 = non-terminal atoms bonded to 4 other atoms (unbonded electrons of O, NH, N, S, are considered a bond) SP2 = non-terminal atoms singly bonded to two other atoms and doubly bonded to a 3rd atom Rings = # independent rings HBN = is the hydrogen bond number as a function of the number of OH, COOH, and NH2 groups

MW

NH.33COOHOHHBN 2++

=

14

15

A more complicated method: From Zhao, H.; Li, P.; Yalkowski, H.; Predicting the Entropy of boiling for Organic Compounds, J. Chem. Inf. Comput. Sci, 39,1112-1116, 1999 ∆∆Sb= 84.53 – 11σσ +.35ττ + 0.05ωω2 + ΣΣΧΧ i

where:

ΧΧ i = the contribution of group i to the Entropy of boiling

ω ω = the molecular planarity number, or the # of non-hydrogen atoms of a molecule that are restricted to a single plane; methane and ethane have values of 1 and 2; other alkanes, 3; butadiene, benzenes, styrene, naphthalene, and anthracene are 4,6,8,10,14

ττ measures the conformational freedom or flexibility ability of atoms in a molecule to rotate about single bonds τ= SP3 + 0.5(SP2) +0.5 (ring) –1

16

σ = symmetry number; the number of identical images that can be produced by a rigid rotation of a hydrogen suppressed molecule; always greater than one; toluene and o-xylene = 2, chloroform and methanol =3, p-xylene and naphthalene = 4, etc

17

18

Boiling points can be estimated based on chemical structure (Joback, 1984)

Tb= 198 + Σ ∆Tb

∆T (oK) -CH3 = 23.58 K -Cl = 38.13 -NH2 = 73.23 C=O = 76.75 CbenzH- = 26.73 Joback obs (K) (K) acetonitrile 347 355 acetone 322 329 benzene 358 353 amino benzene 435 457 benzoic acid 532 522 toluene 386 384 pentane 314 309 methyl amine 295 267 trichlorethylene 361 360 phenanthrene 598 613

19

Stein, S.E., Brown, R.L. Estimating Normal boiling Points from Group Contributions, J.Chem. Inf Comput. Sci, 34, 581-587, 1994

They start with Tb= 198 + Σ ∆Σ ∆Tb and go to 4426 experimental boiling points in Aldrich And fit the residuals (Tb obs-Tb calcd)

20

Tb= 198 + Σ ∆Σ ∆Tb Tb(corr) = Tb- 94.84+ 0.5577Tb- 0.0007705Tb

2 T b< 700 K Tb(corr) = Tb+282.7-0.5209Tb

Tb>700K

21

Estimating Vapor Pressures

2RT

H

dTplnd vap

* ∆=

To estimate the vapor pressure at a temp lower then the boiling temp of the liquid we need to estimate ∆Hvap at lower temperatures. Assume that ∆Hvap is directly proportional to temp and that ∆Hvap can be related to a constant the heat capacity of vaporization ∆Cp Tb where ∆vapH/∆T = ∆Cp Tb

∆vapHT = ∆vapH Tb + ∆CpTb(T-TTb)

)T

bT(ln

R

pTbÄC)

T

bT(1

R

bpTÄC)

T

1

bT

1(

R

bTHvap

Ä*iLPln −−−−=

at the boiling point ∆vapH Tb= Tb ∆vapS Tb

)()((TT

lnR

C)

TT

R

C

R

Spln bpTbbpTTvap* bb

iL

∆−−

∆−

∆= 1

22

)())((TT

lnR

C

TT

R

C

R

Spln bpTbbpTTvap*

iLbb

∆−−

∆−

∆= 1

for many organic compounds ∆Cp Tb/ ∆vapS Tb ranges from -0.6 to -1 so substituting ∆Cp Tb= -0.8 vap∆S Tb

)]()([TT

ln.TT

.R

S*pln bbvapT

iLb 80181 +−

∆=

if we substitute ∆Svap Tb= 88J mol-1 K-1 and R =8.31 Jmol-1 K-1

)]()( ln.ln *

T

T

T

Tp bb

iL 58119 +−=

when using ∆Cp Tb/ ∆Svap Tb = -O.8, low boiling compounds (100oC) are estimated to with in 5%, but high boilers may be a factor of two off If the influence of van der Waal forces (Kistiakowky)and polar and hydrogen bonding effects (Fishtine’s correction factors) are applied ∆∆vapS Tb= Kf(36.6 +8.31 ln Tb)

23

)]()([ ln..)ln.(ln *

T

T

T

TbTfKp bb

iL 8018144 −−+=−

If we go back to:

)())((TT

lnR

C

TT

R

C

R

Spln bpTbbpTTvap*

iLbb

∆−−

∆−

∆= 1

∆∆vapSi (Tb) = 86.0+ 0.04 ττ + 1421 HBN and Mydral and Yalkowsky suggest that

∆∆vapCpi (Tb) = -90 +2.1ττ in J mol-1K-1

τ = Σ(SP3 +0.5 SP2 +0.5 ring) –1

TbT

HBNiLpT

Tb

ln)..(

))(177.21.2(*ln

τ

τ

250810

30 1

++

++− −=

24

MW

NH.33COOHOHHBN 2++

=

25

26

A vapor pressure calculation for the liquid vapor for anthracene

)]()( ln.ln *

T

T

T

Tp bb

iL 58119 +−=

Tb= 198 + Σ ∆Tb ; for anthracene {C14H18} C14H18 Has 10 =CH- carbons at 26.73oK/carbon

And 4, =C< , carbons 31.01OK/carbon

Tb= 589K; CRC = 613K

At 298K, ln p* = -12.76; p = 2.87 x10-6atm =

and p*iL = 0.0022 torr What do we get with the real boiling point of 613K ? ln p*iL = 8.7 x10-7atm

27

Solid vapor pressures For a solid to go directly to the gas phase (sublimation),we can say that the free energy of sublimation(∆∆subGi) has to be the sum of the free energy need to go from the solid to the liquid (∆∆fusGi) + the free energy of going from a liquid to a gas (∆∆vapGi) ∆∆subGi = ∆∆fusGi + ∆∆vapGi

also ∆∆subHi = ∆∆fusHi + ∆∆vapHi and ∆∆subSi = ∆∆fusSi + ∆∆vapSi

so

∆∆fusGi = ∆∆subGi - ∆∆vapGi

for both ∆∆vapGi and ∆∆subGi

the change in free energy in going from one temperature to another, like say ambient to the melting point ∆∆phase j Gi = RT ln { pref / p*ij } so in ∆∆fusGi = ∆∆subGi - ∆∆vapGi

∆∆fusGi = RTln { pis

melt / p*is } - RTln { piLmelt / p*iL }

∆∆fusGi = RTln { p*iL / p*is }

28

this suggests a relationship between p*iL and p*is

29

Solid Vapor Pressures ∆subH = ∆fusH + ∆vapH

∆ fusH (s) ∆fusH= Tm ∆fusS

∆fusS

∆fusH/ Tm = ∆fusS = const?

T ∆Hsub = ∆Hvap+ Tm ∆Sfus

RH

Td

plnd * ∆−=1

It can be shown that

amb

ambmfusiSiL T

TT

R

SPp

)()(lnln ** −∆

+=

if ∆S= const = 56.4 J mol-1K-1 and R=8.31 J mol-1K-1, ∆Sfus /R= 6.78 see if you can derive this

30

What is the solid vapor pressure for anthracene? Using the correct boiling point we determined the liquid vapor pressure to be 8.71x10-7 atmospheres

amb

ambmfusiSiL T

TT

R

Spp

)()(lnln ** −∆

+=

if ∆S= const = 56.4 J mol-1K-1 and R=8.31 J mol-1K-1, ∆Sfus /R= 6.78 ln 8.71x10-7

= ln p*iS+ 6.78 (490.65-298)/298

-13.95 – 4.38 = ln p*iS

7.8x10-9= p*iS

31

Myrdal and Yalkowski also suggest that a reasonable estimate of ∆∆ fus Si(Tm) is

∆fus Si(Tm) + 56.5+ 9.2 ττ -19.2 log σσ) in J mol-1K-1

substitution in to

amb

ambmfusiSiL T

TT

R

SPp

)()(lnln ** −∆

+= gives

amb

ambm

iS

iL

T

TT

Rp

p )()log...(ln

*

* −−+−=

στ 321186

32

Measuring solid vapor pressures

33

Using Sonnefeld et al, what is the sold vapor pressure for anthracene at 289K log10 p

*iS = -A / T + B; p*

iS is in pascals 101,325 pascals = 1atm A= 4791.87 B= 12.977 log10 p

*iS = -4791.87 / T + 12.977

log10 p

*iS = -16.0801 + 12.977 = -3.1031

Po = 7.88 x10-4 pascals p*

iS = 7.88 x10-4 /101,325 = 7.8x10-9 atm

34

A chromatographic Method for measuring liquid vapor pressures

35

A chromatographic Method for measuring liquid vapor pressures A B Hamilton,J.Chrom.195, 75-83,1980 Hinckley et al., J. Chem. Eng. Data, 1990; Yamasaki, 1986 to t1 t2

retention time A = t1- to=t’1

retention volume (Vr) = column flow x the retention time and

t’1/t’2= VrA / VrB and Vr varies inversely with the vapor pressure of a compound VrA / VrB = p*BL/p*AL eq1

rA

rBALBL V

Vpp lnlnln ** −= eq 2

36

)T

(dR

Hplnd Bvap*

BL1∆

−= eq 3

)T

(dR

Hplnd Avap*

AL1∆

−= ; eq 4

dividing eq 4 into 3 and integrating

Avap

Bvap*AL

*BL H

Hplnpln

∆

∆= + const eq 5

we said that

rA

rB*AL

*BL V

Vlnplnpln −= eq 2

combining eqs. 2 and 5

const)H

H(pln

V

Vln

Avap

Bvap*AL

rA

rB−

∆

∆−= 1 eq 6

1.If we can know or can calculate the vapor pressure p*AL at different temperatures

37

2. And run our GC isothermally at these temperatures and obtain of t’2/t’1 and hence VrB / VrA 3.we can plot ln VrB / VrA vs p*AL , get the slope and intercept, and plug into eq 5 to obtain p*BL

38

Using vapor pressure and activity coefficients to estimate organic gas-particle partitioning (Pankow, Atmos. Environ., 1994) Gas Atoxic + liquid particle àà particle Atoxic

+liquid particle

Kip = Aipart / (Aigas xTSP);. TSP has units of ug/m3

Aigas and Aipart have units of ng/m3

log Kip= -log p*iL + const based on solid-gas

partitioning

pi = ΧΧ γ γ p*iL (in atmospheres)

pi V = ni RT/760 = [Aigas] RT/760; (mmHg)

[Aigas] = [PAHigas] = ni /V

Aigas Aipart

39

[PAHigas] = ΧΧ γγ p*iL MWi x 109/( 760 RT)

Let’s look at the mole fraction

ΧΧ i = = moles in the particle phase of i divided by total moles in the particle phase Usually we measure ng/m3 in the particle phase of compound i [Aipart] = [PAHipart] The number of moles in the particle phase is: iMoles = [PAHipart]/ {MWi 109 } = moles/m3

We usually measure TSP as an indicator of total particle mass and the amt. of liquid in the particle phase = TSP x fom The average number of moles in the particle phase requires that we assume an average molecular weight for the organic material in the particle phase, MWavg

40

total liquid moles ofTSP( µµg/m3) = fomTSP/ {MWavg 106 }

we said before iMoles = [Partipart]/ {MWi 109 } = moles/m3 ΧΧ i = = iMoles/ total moles =

[PartiPAH] MWavg / {TSP MWi 103} we also said before that

[PAHigas] = Χi γγ p*iL MWi x 109/( 760 RT)

and Kip = PAHipart / (PAHigas xTSP) Kip = 760 RT fomx10-6/{p*

iLtorr MWavg};

p*

iL here is in torr

41

ln p* =

log p*iL-torr

log

K p

slope = - 1

pyrene

phenanthreneee

BaP

log

K p

slope = - 1

pyrene

BaP